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If $B=S^{-1}AS$, thenso the characteristic polynomials, eigenvalues, and algebraic multiplicities agree. The converse is false: $I_2$ and $\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)$ have the same characteristic polynomial, but the first is diagonalisable and the second is not, so they cannot be similar.

The characteristic polynomial isFor $\lambda =2$, the eigenspace is spanned only by $(1,0,0{)}^T$, so its geometric multiplicity is one although its algebraic multiplicity is two. Hence $M$ is not diagonalisable.

The signed displacement of the sphere centre from the plane is $g\cdot n-p$. Orthogonal projection gives the circle centrePythagoras givesA real circle exists when $|p-g\cdot n|\le A$: equality gives tangency and radius zero, while strict inequality gives a genuine circle.

The cross product of the plane normals is proportional to $(1,-1,-1)$, soThe point on the intersection perpendicular to $m$ iswhich satisfies both plane equations and $u\cdot m=0$. Hence the line is $r\times m=u\times m$.

The equally inclined line has direction $d=(1,1,1)/\sqrt{3}$. The distance between the two skew lines is

No. TakeBoth functions are nonconstant on every interval and nondifferentiable at zero, but $g(f(x))=2x$ is differentiable.

No. Take $f(x)=|x|$, $a=0$, and $g(y)=y^2$. Then $f$ is not differentiable at zero, but $g\circ f=x^2$ is differentiable there.

No. Take $f(x)=x^2$, $a=0$, and $g(y)=|y|$. The function $g$ is not differentiable at $f(0)=0$, but $g(f(x))=x^2$ is differentiable. Neither function is constant on an interval.

Apply part (b) to the normal matrix $C=D-\lambda I$. Since $(D-\lambda I)e=0$, one hasThus $e$ is an eigenvector of $D^{†}$ with eigenvalue $\overline{\lambda }$.

Normality givesTherefore one norm vanishes exactly when the other does.

For $B=A^{-1}{A}^{†}$,Multiplying $B^{†}B=I$ on the left by $A^{-1}$ and on the right by $(A^{†}{)}^{-1}$ shows that it is equivalent to $(A^{†}{)}^{-1}{A}^{-1}=A^{-1}(A^{†}{)}^{-1}$. Taking inverses gives $A^{†}A=A{A}^{†}$. Thus $B$ is unitary exactly when $A$ is a normal matrix.

Sincethe Inverse Fourier transforms of phase factors giveand

From $z=\tan w=-i(e^{2iw}-1)/(e^{2iw}+1)$ one obtainsFor $z=(2\sqrt{3}-3i)/7$ the quotient is $1+i\sqrt{3}=2e^{i\pi /3}$. Its principal logarithm is $\log 2+i\pi /3$, so the principal value is

The Fourier transform of a derivative givesSince $p^{\prime }=g$,An arbitrary additive constant in $p$ affects only the zero-frequency Dirac delta function when transforms are interpreted as distributions.

The values of $\log i$ are $i(\pi /2+2\pi n)$. HenceThese are infinitely many distinct positive real numbers.

For the stated angular-frequency Fourier transform convention, the convolution theorem givesThe Fourier inversion theorem and the assumed absolute integrability therefore imply

Since $|-i|=1$ and the principal argument is $-\pi /2$,All logarithms are $-\pi i/2+2\pi in$, $n\in \mathbb{Z}$.

The continuous approximation of a Riemann-integrable function follows here by choosing partitions $P_n$ with upper-minus-lower sum below $1/n$. On each subinterval choose the midpoint of the infimum and supremum to form a step function $s_n$. ThenReplace each of the finitely many jumps of $s_n$ by a linear transition on intervals of sufficiently small total length, obtaining a continuous ${\varphi }_n$ with $\int |s_n-{\varphi }_n|<1/n$. Hence $\int |g-{\varphi }_n|<2/n$, and uniformly for every subinterval $[\alpha ,\beta ]$,

Changing variables givesThe average of a continuous function over $[y,y+h]$ tends to $f(y)$ as $h\to 0$. The two terms therefore tend to $f(x)$ and $f(a)$, proving the stated limit.

For a bounded function, the lower and upper integrals are the supremum of lower Darboux sums and infimum of upper Darboux sums. It is Riemann integrable when these agree.

Here $u(x)=1/x-\lfloor 1/x\rfloor$ for $x>0$, so $0\le u<1$. On $[\delta ,1]$ it has only finitely many discontinuities and is piecewise continuous, hence integrable. On $[0,\delta ]$ its upper-minus-lower contribution is at most $\delta$. Taking $\delta \downarrow 0$ proves integrability on $[0,1]$.

No. The Darboux theorem says that every derivative has the intermediate-value property. Choose $ϵ<1/3$. In a deleted neighborhood of $a$, the assumed limit puts $\varphi (x)$ within $ϵ$ of $\varphi (a)+1$. Darboux's theorem applied between $a$ and any such $x$ would require values near $\varphi (a)+1/2$, which that neighborhood excludes. This contradiction shows that $\varphi$ cannot be a derivative.